1.

Develop the notions of work and show that it leads to work-energy theorem . State the condition under which a force does no work .

Answer»

Solution :WORK : The product of component of force in the direction of displacement and the magnitude of displacementis called work .
`W=bar(F).bar(S)`
When `bar(F)andbar(S) " are parallel "W=|bar(F)|xx|bar(S)|`
When`bar(F)andbar(S)` has some angle`theta` between them
`W=bar(F).bar(S)costheta`

Kinetic energy : Energy possessed by a moving body is called kinetic energy (k)
The kinetic energy of an OBJECT is a measure of the work that an object can do by the virtue of its motion .
Kinetic energy can be measured with equation `K=(1)/(2)mv^(2)`
Ex : All moving bodies contain kinetic energy .
Work energy theorem (For variable force ) : work done by a variable force is always equal to the change in kinetic energy of the body .
Work done `W=(1)/(2)mV^(2)-(1)/(2)mV_(0)^(2)=K_(f)-K_(i)`
Proof : Kinetic energy of a body `K=(1)/(2)mv^(2)`
Time rate of change of kinetic energy is
`(DK)/(dt)=(d)/(dt)((1)/(2)mv^(2))=(1)/(2)m.2v.(dv)/(dt)=mv.(dv)/(dt)`
But `(dv)/(dt)=a`
`therefore(dk)/(dt)=mav " but"ma="Force"(F)`
`therefore(dk)/(dt)=Fv=F.(dx)/(dt)RARR(dk)/(dt)=F.(dx)/(dt) " or "dk=f.dx`
When force is conservation force F = F (x)
`therefore` On INTEGRATION over initial position`(X_(1))`and final position `(X_(2))`
`int_(i)^(f)dk=int_(x_(1))^(x_(2))F(X)dx " but "int_(x_(1))^(x_(2))F(X)dx=` work done By variable force W
`thereforeW=int_(i)^(f)rArrW=K_(f)-K_(i)`
I.e ., work done by a conservative force is equal to chage in kinetc energy of the body .
Coondition for force not to do any work .
When Force `(bar(F))` and displacement `(bar(S))` are perpendicular work done is zero . , i.e., when `theta=90^(@)" then " W = bar(F). bar(S)=0`


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